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5.3 (a) Looking into the base of Q 1 we see an equivalent resistance of r π1 , so we can draw the following equivalent circuit for finding R in : R in R 1 R 2 r π1 R in…
3.1 (a) IX = VX R1 0 VX < 0 VX > 0 IX VX (V) Slope = 1/R1 3.2 IX = Plotting IX (t), we have VX R1 0 VX < 0 VX > 0 V0 IX (t) for VB = 1 V (Solid) 0 0 −V0 /R1…
4.4 According to Equation (4.8), we have IC = AE qDn n2 VBE /VT i −1 e NB WB 1 ∝ WB We can see that if WB increases by a factor of two, then IC decreases by a factor…
3.1 (a) IX = { VX R1 VX < 0 0 VX > 0 VX (V) IX Slope = 1/R1 3.2 IX = { VX R1 VX < 0 0 VX > 0 Plotting IX(t), we have 0 −V0/R1I X (t ) fo r V B = 1 V (S ol id…
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BASIC ELECTRONICS ASSIGNMENT-II TRANSISTORS: 1. A BJT has alpha 0.99 and collector-to-base reverse sat current 3μA. If emitter current is 10mA, calculate the collector and…
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